In this proof, we will show that if a^2b^2c^2 are in AP (Arithmetic Progression), then a/b, b/c, c/a, ac/b, and bc/a are also in AP.


Let's assume that a^2b^2c^2 are in AP. This means that:

2b^2 = a^2 + c^2
2a^2 = b^2 + c^2
2c^2 = a^2 + b^2

We can rearrange these equations to get:

a^2 - 2b^2 + c^2 = 0
b^2 - 2a^2 + c^2 = 0
a^2 - 2c^2 + b^2 = 0

We can factorize these equations to get:

(a - b)(a + b - 2c) = 0
(b - c)(b + c - 2a) = 0
(a - c)(a + c - 2b) = 0

From these factorizations, we can see that a = b, b = c, or a = c. However, we know that a, b, and c are distinct, so none of these equalities hold.

Therefore, we can divide each equation by the corresponding factor to get:

a/b + b/c = 2(a^2b^2c^2)/(abc)^2 = 2c/a
b/c + c/a = 2(a^2b^2c^2)/(abc)^2 = 2b/a
c/a + a/b = 2(a^2b^2c^2)/(abc)^2 = 2c/b

These equations show that a/b, b/c, and c/a are in AP. Similarly, we can show that ac/b and bc/a are in AP by dividing the equations:

ac/b + bc/a = 2(a^2b^2c^2)/(abc)^2 = 2c^2/b^2
bc/a + ac/b = 2(a^2b^2c^2)/(abc)^2 = 2a^2/b^2

Therefore, we have shown that a/b, b/c, c/a, ac/b, and bc/a are in AP if a^2b^2c^2 are in AP.


In conclusion, we have proved that if a^2b^2c^2 are in AP, then a/b, b/c, c/a, ac/b, and bc/a are also in AP. This is an important result in mathematics and has numerous applications in different fields.